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Calculadora de sistemas de equações lineares

Solve 2 to 5 simultaneous linear equations exactly by Gauss–Jordan elimination, with every row operation shown, and detect no-solution or infinite cases.

Atualizado Exemplos verificados: 5

One equation per line: the coefficients, then the constant. 2 1 -1 8 means 2x + y − z = 8. A | or = before the constant is fine; fractions like 1/2 work.
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Solution
x = 2, y = 3, z = −1
Solution: x = 2, y = 3, z = −1
Type of system
Unique solution
x₁ (x)
2
x₂ (y)
3
x₃ (z)
−1
Determinant of the coefficients
−1
Condition number (∞-norm)
60

The 3 equations meet at exactly one point: x = 2, y = 3, z = −1. The determinant is −1, which is non-zero, so the solution is unique.

Solution values

x2y3z−1
Como é calculado S

  1. Augmented matrix

    [A∣b]=[21−18−3−12−11−212−3]\left[A \mid \mathbf b\right] = \left[\begin{array}{ccc|c}2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3\end{array}\right]

    Unknowns: x, y, z.

  2. Column 1: make the pivot 1 and clear the rest of the column

    R1←12 R1, R2←R2+3R1, R3←R3+2R1  ⟶  [112−1240121210215]R_{1} \leftarrow \frac{1}{2}\,R_{1},\ R_{2} \leftarrow R_{2} + 3R_{1},\ R_{3} \leftarrow R_{3} + 2R_{1} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5\end{array}\right]

  3. Column 2: make the pivot 1 and clear the rest of the column

    R2←2 R2, R1←R1−12R2, R3←R3−2R2  ⟶  [10−13011200−11]R_{2} \leftarrow 2\,R_{2},\ R_{1} \leftarrow R_{1} - \frac{1}{2}R_{2},\ R_{3} \leftarrow R_{3} - 2R_{2} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1\end{array}\right]

  4. Column 3: make the pivot 1 and clear the rest of the column

    R3←−R3, R1←R1+R3, R2←R2−R3  ⟶  [10020103001−1]R_{3} \leftarrow -R_{3},\ R_{1} \leftarrow R_{1} + R_{3},\ R_{2} \leftarrow R_{2} - R_{3} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1\end{array}\right]

  5. Read off the solution

    x=2,y=3,z=−1x = 2,\quad y = 3,\quad z = -1

Sobre Calculadora de sistemas de equações lineares

A system of linear equations is written as an augmented matrix [A | b], one row per equation. Gauss–Jordan elimination scales each pivot row so the pivot is 1 and subtracts multiples of it from every other row until the coefficients form the identity matrix, leaving the solution in the last column. The arithmetic uses exact fractions, so 2x + 3y = 7, 4x − y = 1 gives x = 5/7 and y = 13/7 rather than rounded decimals.

Simultaneous equations come up in circuit analysis, mixture and pricing problems, and algebra courses. The default is the 3 × 3 example from Wikipedia's Gaussian elimination article, 2x + y − z = 8, −3x − y + 2z = −11, −2x + y + 2z = −3, with the unique solution x = 2, y = 3, z = −1.

By the Rouché–Capelli theorem the system has one solution when the rank of A equals the number of unknowns, infinitely many when A and [A | b] share a smaller rank, and none when their ranks differ.

Exemplos resolvidos

Wikipedia 3×3 example

Equations as rows of numbers
2 1 -1 8 -3 -1 2 -11 -2 1 2 -3
Solution
x = 2, y = 3, z = −1
x₁ (x)
2
x₂ (y)
3
x₃ (z)
-1
Determinant of the coefficients
-1
Condition number (∞-norm)
60

Fonte de verificação: Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10

Fractional answer: 2x + 3y = 7, 4x − y = 1

Equations as rows of numbers
2 3 7 4 -1 1
Solution
x = 5/7, y = 13/7
x₁ (x)
0.7142857143
Determinant of the coefficients
-14

Fonte de verificação: Python fractions: x = 10/14, y = 4x − 1 = 13/7

Inconsistent: x + y = 1, 2x + 2y = 3

Equations as rows of numbers
1 1 1 2 2 3
Type of system
No solution — the equations are inconsistent
Solution
No solution

Fonte de verificação: 2 × (first) gives 2x + 2y = 2 ≠ 3

Dependent equations (edge case)

Equations as rows of numbers
1 1 1 6 2 2 2 12 1 -1 0 0
Type of system
Infinitely many solutions
Solution
x = 3 − t/2, y = 3 − t/2, z = t
Determinant of the coefficients
0

Fonte de verificação: Row 2 = 2 × row 1; x = y and 2y + z = 6 (hand elimination)

Perguntas

How do you solve a system of three equations with three unknowns?

Eliminate one variable at a time. Gauss–Jordan elimination writes the equations as a matrix, turns the first coefficient into 1, subtracts multiples of that row to clear x from the other equations, then repeats for y and z. For the default system three rounds of row operations leave x = 2, y = 3, z = −1; substituting into 2x + y − z = 8 gives 4 + 3 + 1 = 8.

How can you tell if a system has no solution or infinitely many?

Row-reduce it and look at rows whose coefficients all become 0. A row reading 0 = 1, or 0 equal to any non-zero number, means no solution: x + y = 1 and 2x + 2y = 3 are parallel lines. A row reading 0 = 0 means one equation repeats another, leaving a free variable and infinitely many solutions. With as many equations as unknowns, a determinant of 0 signals one of these two cases.

What is Cramer's rule?

Cramer's rule gives each unknown as a ratio of determinants: x = det(Aₓ)/det(A), where Aₓ is A with its x column replaced by the constants. For 2x + 3y = 7, 4x − y = 1, det(A) = 2 × (−1) − 3 × 4 = −14 and det(Aₓ) = 7 × (−1) − 3 × 1 = −10, so x = 10/14 = 5/7. It needs det(A) ≠ 0 and much more arithmetic than elimination beyond 3 × 3.

What is the difference between Gaussian and Gauss–Jordan elimination?

Gaussian elimination stops at row echelon form, a triangular matrix, and finds the unknowns by back-substitution from the last equation upward. Gauss–Jordan continues until the coefficients form the identity matrix, so each row reads off one unknown directly. Gauss–Jordan takes about n³ arithmetic operations for n equations against about 2n³/3, in exchange for skipping the substitution step.

What does the condition number of a system mean?

It bounds how much a small relative change in the coefficients or constants can grow in the solution. The ∞-norm version is κ = ‖A‖∞ × ‖A⁻¹‖∞; for the default system ‖A‖∞ = 6 and ‖A⁻¹‖∞ = 10, so κ = 60. As a rule of thumb about log₁₀ κ significant digits are lost, here about 2, and a value near 10¹² means the equations are nearly dependent.

Qual é a precisão de “Calculadora de sistemas de equações lineares”?

A precisão depende dos dados inseridos e das hipóteses do método. O cálculo decimal usa 50 algarismos significativos, mas estimativas, métodos numéricos e dados de origem podem ter menor precisão; o arredondamento exibido não elimina essas limitações. Exemplos resolvidos verificados com fontes independentes: 5. Por exemplo, “Wikipedia 3×3 example” é verificado com Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10.

De onde vem o método?

Wikipedia — Gaussian elimination (worked 3×3 example); Wolfram MathWorld — Rouché–Capelli (Kronecker–Capelli) theorem: consistency by rank.

Sobre esta calculadora

[A∣b]→row operations[I∣x]κ∞(A)=∥A∥∞ ∥A−1∥∞\begin{gathered} [A \mid \mathbf b] \xrightarrow{\text{row operations}} [I \mid \mathbf x] \\[10pt] \kappa_\infty(A) = \|A\|_\infty \, \|A^{-1}\|_\infty \end{gathered}

Fontes

  1. Wikipedia — Gaussian elimination (worked 3×3 example)
  2. Wolfram MathWorld — Rouché–Capelli (Kronecker–Capelli) theorem: consistency by rank

Verificado com as referências

Esta calculadora inclui 5 exemplos resolvidos com respostas de fontes independentes. Eles fazem parte do conjunto de testes e você também pode executá-los aqui.

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